Work Done in Reversible Cyclic Process for Gas A

To find the work done in the reverse process, we apply the First Law of Thermodynamics, which states:

$\Delta U = q + w$

where:

• $\Delta U$ is the change in internal energy,
• $q$ is the heat supplied to the system,
• $w$ is the work done on the system (using the IUPAC convention: $w = -w_{\text{by}}$).

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Step 1: Process $X \to Y$

For the process from state $X$ to state $Y$:

• Heat absorbed by the gas, $q_1 = +10\text{ J}$
• Work done by the gas, $w_{\text{by}, 1} = +18\text{ J} \implies w_1 = -18\text{ J}$

Using the First Law of Thermodynamics: $\Delta U_{X \to Y} = q_1 + w_1 = 10\text{ J} - 18\text{ J} = -8\text{ J}$

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Step 2: Process $Y \to X$

Since internal energy $U$ is a state function, the change in internal energy for the reverse path $Y \to X$ must be equal in magnitude and opposite in sign to that of $X \to Y$:

$\Delta U_{Y \to X} = -\Delta U_{X \to Y} = -(-8\text{ J}) = +8\text{ J}$

For this reverse process:

• Heat evolved during the process, $q_2 = -6\text{ J}$

Applying the First Law of Thermodynamics for $Y \to X$: $\Delta U_{Y \to X} = q_2 + w_2$ $8\text{ J} = -6\text{ J} + w_2$ $w_2 = 8\text{ J} + 6\text{ J} = +14\text{ J}$

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Conclusion

A positive value of $w_2 = +14\text{ J}$ signifies that $14\text{ J}$ of work is done on the gas 'A' by the surroundings.

Hence, the correct option is D.